|
| 1 | +import LinkedList from '../link-list/LinkedList' |
| 2 | + |
| 3 | + |
| 4 | +// 积水问题,使用单调栈的思想 |
| 5 | +class Solution42 { |
| 6 | + trap(height) { |
| 7 | + const stack = new LinkedList(); |
| 8 | + // 最少也需要3个柱子才能积水,否则直接返回0 |
| 9 | + if (height.length < 3) return 0; |
| 10 | + |
| 11 | + let res = 0; |
| 12 | + for (let i = 0; i < height.length; i++) { |
| 13 | + // 当栈不为空,且不满足递减关系 弹栈并记录弹出元素下标,用于计算积水高度 |
| 14 | + while (!stack.isEmpty() && height[i] > height[stack.getLast()]) { |
| 15 | + let temp = stack.deleteTail(); |
| 16 | + // 当有重复连续值时都弹出 |
| 17 | + while (!stack.isEmpty() && height[temp] == height[stack.getLast()]) { |
| 18 | + stack.deleteTail(); |
| 19 | + } |
| 20 | + //计算积水深度 |
| 21 | + if (!stack.isEmpty()) { |
| 22 | + // 计算宽度 |
| 23 | + let width = i - stack.getLast() - 1; |
| 24 | + // 计算高度 |
| 25 | + let high = Math.min(height[i] - height[temp], height[stack.getLast()] - height[temp]); |
| 26 | + // System.out.println(i + " " + width + " " + high + " " + temp); |
| 27 | + res += high * width; |
| 28 | + } |
| 29 | + } |
| 30 | + stack.append(i); |
| 31 | + // System.out.println(); |
| 32 | + } |
| 33 | + |
| 34 | + return res; |
| 35 | + } |
| 36 | +} |
| 37 | + |
| 38 | +const s = new Solution42() |
| 39 | +const result= s.trap([4,2,0,3,2,5]) |
| 40 | +console.log('water-problem',result) |
| 41 | +// // 按层计算,负责度较高可能会超时 |
| 42 | +// class Solution42_byLayer { |
| 43 | +// public let trap(let[] height) { |
| 44 | +// let res = 0, max = 0; |
| 45 | +// for (let i = 0; i < height.length; i++) { |
| 46 | +// max = Math.max(height[i], max); |
| 47 | +// } |
| 48 | + |
| 49 | +// // 按层计算 |
| 50 | +// for (let i = 1; i <= max; i++) { |
| 51 | +// boolean flag = false; |
| 52 | +// let ceng = 0; |
| 53 | +// for (let j = 0; j < height.length; j++) { |
| 54 | +// // 如果有小于i的且左边有>=i的则此处可以存水 |
| 55 | +// if (flag && height[j] < i) { |
| 56 | +// ceng++; |
| 57 | +// } |
| 58 | +// // 可挡住水的墙把flag设置为True 把ceng加到总数上再置零计算下一个积水处 |
| 59 | +// if (height[j] >= i) { |
| 60 | +// flag = true; |
| 61 | +// res += ceng; |
| 62 | +// ceng = 0; |
| 63 | +// } |
| 64 | +// } |
| 65 | +// } |
| 66 | + |
| 67 | +// return res; |
| 68 | +// } |
| 69 | +// } |
| 70 | + |
| 71 | + |
| 72 | +// // 按列计算,第一列和最后一列不可能存水直接跳过 1~height.length-1 |
| 73 | +// class Solution42_byClo { |
| 74 | +// public let trap(let[] height) { |
| 75 | +// let res = 0, max = 0; |
| 76 | + |
| 77 | +// // 第一列和最后一列不可能存水直接跳过 1~height.length-1 |
| 78 | +// for (let i = 1; i < height.length - 1; i++) { |
| 79 | +// let left = 0, right = 0, min = 0; |
| 80 | + |
| 81 | +// // 计算左边最大值 |
| 82 | +// for (let j = i - 1; j >= 0; j--) { |
| 83 | +// left = Math.max(left, height[j]); |
| 84 | +// } |
| 85 | + |
| 86 | +// //计算右侧最大值 |
| 87 | +// for (let j = i + 1; j < height.length; j++) { |
| 88 | +// right = Math.max(right, height[j]); |
| 89 | +// } |
| 90 | + |
| 91 | +// min = Math.min(right, left); |
| 92 | + |
| 93 | +// // i列积水深度 |
| 94 | +// let deep = min - height[i]; |
| 95 | +// if (deep > 0) { |
| 96 | +// res += deep; |
| 97 | +// } |
| 98 | +// } |
| 99 | + |
| 100 | +// return res; |
| 101 | +// } |
| 102 | +// } |
| 103 | + |
| 104 | +// // 双指针法 |
| 105 | +// // 设置两个指针向中间移动,用maxLeft、maxRight分别记录左侧和右侧的最值 |
| 106 | +// // 若maxLeft < maxRight 则以左侧为依据,否则以右侧为依据 |
| 107 | +// // 以左侧为依据时候,当前左指针指向元素height[left]>=maxLeft 则该地方不能积水,left++向右移动。否则res += maxLeft - height[left] |
| 108 | +// // 以右侧为依据时候,当前右指针指向元素height[right]>=maxRight 则改地方不能积水。否则res += maxRight - height[left] |
| 109 | +// class Solution42_byDoublePointer { |
| 110 | +// public let trap(let[] height) { |
| 111 | +// let res = 0, maxLeft = height[0], max_Right = height[height.length - 1], left = 0, right = height.length - 1; |
| 112 | +// while (left <= right) { |
| 113 | +// // 如果左侧比较小,则以左侧为准,判断height[left]处是否积水 |
| 114 | +// if (maxLeft < max_Right) { |
| 115 | +// // 如果左侧最大值比当前左指针位置大则可以积水 |
| 116 | +// if (height[left] < maxLeft) { |
| 117 | +// res += maxLeft - height[left]; |
| 118 | +// } else { // 否则该位置不能积水,因为其比左侧最大高度低 不能积水 |
| 119 | +// maxLeft = Math.max(height[left], maxLeft); |
| 120 | + |
| 121 | +// } |
| 122 | +// left++; |
| 123 | +// } else { // 此时右侧高度比较小,判断右侧位置情况 |
| 124 | +// // 此时 height[right] 比较小可以积水 |
| 125 | +// if (height[right] < max_Right) { |
| 126 | +// res += max_Right - height[right]; |
| 127 | +// } else { // 此时不能积水 |
| 128 | +// max_Right = Math.max(max_Right, height[right]); |
| 129 | + |
| 130 | +// } |
| 131 | +// right--; // 指针左移动 |
| 132 | +// } |
| 133 | + |
| 134 | +// } |
| 135 | +// return res; |
| 136 | +// } |
| 137 | +// } |
| 138 | + |
| 139 | +// public class LeetCode42 { |
| 140 | +// @Test |
| 141 | +// public void test() { |
| 142 | +// // let[] height = {0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1}; |
| 143 | +// let [] height = { 2, 0, 2}; |
| 144 | +// System.out.println(new Solution42().trap(height)); |
| 145 | +// System.out.println(new Solution42_byLayer().trap(height)); |
| 146 | +// System.out.println(new Solution42_byClo().trap(height)); |
| 147 | +// System.out.println(new Solution42_byDoublePointer().trap(height)); |
| 148 | +// } |
| 149 | +// } |
| 150 | + |
0 commit comments