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Update Leetcode 题解 - 双指针.md
补充三数之和、接雨水、最小覆盖子串，并配图说明接雨水的水位计算。
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CyC2018
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* [5. 归并两个有序数组](#5-归并两个有序数组)
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* [6. 判断链表是否存在环](#6-判断链表是否存在环)
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* [7. 最长子序列](#7-最长子序列)
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* [8. 三数之和](#8-三数之和)
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* [9. 接雨水](#9-接雨水)
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* [10. 最小覆盖子串](#10-最小覆盖子串)
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<!-- GFM-TOC -->
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@@ -290,3 +293,124 @@ private boolean isSubstr(String s, String target) {
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return j == target.length();
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}
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```
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## 8. 三数之和
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15\. 3Sum (Medium)
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[Leetcode](https://leetcode.com/problems/3sum/) / [力扣](https://leetcode.cn/problems/3sum/)
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```text
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Input: nums = [-1, 0, 1, 2, -1, -4]
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Output: [[-1, -1, 2], [-1, 0, 1]]
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```
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题目描述：找出所有和为 0 且不重复的三元组。
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先排序，然后枚举第一个数 nums[i]，在它右边用第 1 题的双指针方法寻找和为 -nums[i] 的两个数。去重的关键是跳过相同的值：i 跳过与前一个数相同的值，找到一组解之后，两个指针也跳过相同的值。时间复杂度为 O(N<sup>2</sup>)。
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```java
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public List<List<Integer>> threeSum(int[] nums) {
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Arrays.sort(nums);
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List<List<Integer>> ret = new ArrayList<>();
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for (int i = 0; i < nums.length - 2; i++) {
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if (i > 0 && nums[i] == nums[i - 1]) {
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continue;
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}
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int l = i + 1, h = nums.length - 1;
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while (l < h) {
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int sum = nums[i] + nums[l] + nums[h];
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if (sum < 0) {
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l++;
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} else if (sum > 0) {
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h--;
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} else {
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ret.add(Arrays.asList(nums[i], nums[l], nums[h]));
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while (l < h && nums[l] == nums[l + 1]) {
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l++;
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}
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while (l < h && nums[h] == nums[h - 1]) {
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h--;
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}
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l++;
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h--;
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}
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}
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}
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return ret;
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}
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```
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## 9. 接雨水
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42\. Trapping Rain Water (Hard)
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[Leetcode](https://leetcode.com/problems/trapping-rain-water/) / [力扣](https://leetcode.cn/problems/trapping-rain-water/)
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```text
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Input: height = [0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1]
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Output: 6
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```
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位置 i 上方能存的水，取决于它左边最高的柱子和右边最高的柱子中较矮的那一个：water[i] = min(leftMax, rightMax) - height[i]。
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用两个指针从两端向中间移动，分别维护 leftMax 和 rightMax。如果 leftMax < rightMax，左指针处的水位只由 leftMax 决定，因为右边已经有一根不低于 rightMax 的柱子，可以直接结算左指针并右移；反之结算右指针并左移。
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<div align="center"> <img src="pics/trapping-rain-water.png" width="560px"/> </div><br>
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```java
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public int trap(int[] height) {
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int l = 0, r = height.length - 1;
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int leftMax = 0, rightMax = 0, ret = 0;
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while (l < r) {
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leftMax = Math.max(leftMax, height[l]);
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rightMax = Math.max(rightMax, height[r]);
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if (leftMax < rightMax) {
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ret += leftMax - height[l++];
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} else {
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ret += rightMax - height[r--];
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}
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}
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return ret;
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}
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```
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## 10. 最小覆盖子串
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76\. Minimum Window Substring (Hard)
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[Leetcode](https://leetcode.com/problems/minimum-window-substring/) / [力扣](https://leetcode.cn/problems/minimum-window-substring/)
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```text
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Input: s = "ADOBECODEBANC", t = "ABC"
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Output: "BANC"
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```
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题目描述：找出 s 中包含 t 所有字符（包括重复字符）的最短子串。
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使用滑动窗口。need[c] 表示窗口还缺几个字符 c，missing 表示一共还缺几个字符。右指针不断扩大窗口，直到 missing 为 0，此时窗口包含了 t 的所有字符；然后移动左指针缩小窗口，在窗口仍然满足条件时更新答案。
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```java
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public String minWindow(String s, String t) {
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int[] need = new int[128];
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for (char c : t.toCharArray()) {
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need[c]++;
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}
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int missing = t.length(), start = 0, minLen = Integer.MAX_VALUE;
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for (int l = 0, r = 0; r < s.length(); r++) {
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if (need[s.charAt(r)]-- > 0) {
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missing--;
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}
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while (missing == 0) {
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if (r - l + 1 < minLen) {
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minLen = r - l + 1;
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start = l;
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}
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if (++need[s.charAt(l++)] > 0) {
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missing++;
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}
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}
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}
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return minLen == Integer.MAX_VALUE ? "" : s.substring(start, start + minLen);
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}
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```
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<rect width="100%" height="100%" fill="#FFFFFF"/>
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<text x="300" y="28" class="title">接雨水</text>
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<text x="78" y="246" class="c sub">0</text>
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<rect x="98" y="190" width="36" height="40" fill="#9E9E9E"/>
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<text x="116" y="246" class="c sub">1</text>
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<rect x="136" y="190" width="36" height="40" fill="#D6EEF8"/>
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<text x="154" y="246" class="c sub">2</text>
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<text x="192" y="246" class="c sub">3</text>
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<rect x="212" y="190" width="36" height="40" fill="#9E9E9E"/>
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<text x="230" y="246" class="c sub">4</text>
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<rect x="288" y="190" width="36" height="40" fill="#9E9E9E"/>
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<text x="306" y="246" class="c sub">6</text>
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<rect x="326" y="110" width="36" height="120" fill="#9E9E9E"/>
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<text x="344" y="246" class="c sub">7</text>
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<rect x="364" y="150" width="36" height="80" fill="#9E9E9E"/>
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<text x="382" y="246" class="c sub">8</text>
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<rect x="402" y="150" width="36" height="40" fill="#D6EEF8"/>
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<rect x="402" y="190" width="36" height="40" fill="#9E9E9E"/>
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<text x="420" y="246" class="c sub">9</text>
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<rect x="440" y="150" width="36" height="80" fill="#9E9E9E"/>
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<text x="458" y="246" class="c sub">10</text>
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<text x="168" y="144" class="e note">leftMax = 2</text>
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<text x="368" y="124" class="note">rightMax = 3</text>
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<line x1="268" y1="86" x2="268" y2="146" class="arrow red"/>
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<text x="208" y="80" class="c">位置 5 存水 min(2, 3) - 0 = 2</text>
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<text x="300" y="278" class="c sub">灰色为柱子，浅蓝色为积水，总共 6</text>
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