标签:ace you not big with integer tween eth rms
Container With Most Water https://www.youtube.com/watch?v=IONgE6QZgGI Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water. Note: You may not slant the container and n is at least 2. ? The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49. Example: Input: [1,8,6,2,5,4,8,3,7] Output: 49 class Solution { public int maxArea(int[] height) { int ans = 0; int left = 0; int right = height.length - 1; while(left < right){ ans = Math.max(ans, Math.min(height[left], height[right]) * (right - left)); if(height[left] < height[right]){ left++; }else{ right--; } } return ans; } } // time : n // height[left] < height[right], if move right towards the inner part. its guarentred that // the area is gonna be smaller, draw a graph and see if makes sense // if move left pointer towards the inner, the area can be larger, samller or equal, we don‘t // really know, it depends on the the second left value, the one we are currently at. // between a choice that the area is def gonna be smaller and a choice that the area might be larger // or smaller or the same, we choose the later // // move two pointers from opposite directions towards the inner is searching for a better area , bigger one
42. Trapping Rain Water Sol1: Time Space class Solution { public int trap(int[] height) { if(height == null || height.length == 0) return 0; int res = 0; int n = height.length; int[] leftMax = new int[n]; int[] rightMax = new int[n]; leftMax[0] = height[0]; rightMax[n - 1] = height[n-1]; // from left to right for(int i = 1; i < n; i++){ leftMax[i] = Math.max(height[i], leftMax[i-1]); } for(int j = n - 2; j >= 0; j--){ rightMax[j] = Math.max(height[j], rightMax[j + 1]); } for(int i = 0; i < n; i++){ res += Math.min(leftMax[i], rightMax[i]) - height[i]; } return res; } } ================ Solution 2 : space O(1) https://www.youtube.com/watch?v=8BHqSdwyODs Idea: two pointers from opposite directions, whose max is smaller, get the result from that side
标签:ace you not big with integer tween eth rms
原文地址:https://www.cnblogs.com/tobeabetterpig/p/9550934.html