基础九 子数组和前缀和

1. 53 Maximum Subarray 找和最大子数组（找最小的话 元素取反求最大就行） - 从前向后 计算sum同时 维持最小的前缀和 - dp dp[i] = nums[i] + (dp[i - 1] > 0 ? dp[i - 1] : 0); dp[i]是已i元素结尾的子数组最大和 2. lt138 Subarray Sum - 题目：找一个和为零的子数组 - 思路：转化为找两个前缀和相同的子数组 中间的数和为零； - 注意：要put(0, -1)；找到的和相同index返回时要加一 因为是要求中间的数 不包括上一个 *3. lt139 Subarray Sum Closest 找一个和最接近零的子数组 转化为找两个前缀和最接近的子数组;注意要添加Pair[0, 0] 计算下表0-2子数组和 用到presum[2]-presum[-1] - Prefix Sum + Sort (Offline Algorithm) Find the subarray sum closet to 0 => find two prefix sum closet to each other => find min difference in sorted prefix sum array Sort (prefixSum, index) pair Time Complexity: O(nlogn) - 另一种解法 Prefix Sum + TreeMap (Online Algorithm) Key: prefix sum Value: index TreeMap is implemented using Red Black Tree (Balanced BST), can be used to find the closet element Ceiling/Floor Time Complexity: O(nlogn) 4. 121 Best Time to Buy and Sell Stock 有一点前缀和的意思 不过注意这里是要求利益的前缀和 所以用前缀和反而没有直接记录之前最小并维持一个全局最大利益来的简单 *5. lt405 Submatrix Sum - 题目：求一个二维数组中某个子矩阵和为零 - 思路：枚举上行下行并在之前做前缀和 O(n^3) - 注意：和之前那个题一样 要put(0, -1)；找到的和相同index返回时要加一 因为是要求中间的数 不包括上一个 6. lt944 Maximum Submatrix - 题目：求一个二维数组中最大子矩阵和 - 思路：枚举上行下行并在之前做前缀和 O(n^3) 7. 304 Range Sum Query 2D - Immutable 8. 308 Range Sum Query 2D - Mutable ####53. Maximum Subarray 找和最大的子数组 ``` class Solution { public int maxSubArrayPrefixSum(int[] nums) { if(nums==null || nums.length==0) return 0; int min = 0; int sum = 0; int result = Integer.MIN_VALUE; for(int i=0; i< n; + (dp[i - 1]> 0 ? dp[i - 1] : 0); max = Math.max(max, dp[i]); } return max; } ``` ####lt138 Subarray Sum ``` public class Solution { /** * @param nums: A list of integers * @return: A list of integers includes the index of the first number and the index of the last number */ public List subarraySum(int[] nums) { // write your code here List list = new ArrayList<>(); Map map = new HashMap<>(); map.put(0, -1); int sum = 0; for(int i=0; i () { public int compare(pair a, pair b) { return a.sum - b.sum; } }); int ans = Integer.MAX_VALUE; for(int i=1; i sums[i].sum-sums[i-1].sum){ ans = sums[i].sum-sums[i-1].sum; result[0] = sums[i-1].index-1; result[1] = sums[i].index-1; Arrays.sort(result); result[0]++; } } return result; } } ``` ####121 Best Time to Buy and Sell Stock ``` class Solution { public int maxProfit(int[] nums) { if(nums==null || nums.length<=1){ return 0; } int max = 0; int min = Integer.MAX_VALUE; for(int i=0; i map = new HashMap<>(); map.put(0, -1); int sum = 0; for(int i=0; i 0 ? sum + sums[bot][i] - sums[top-1][i] : sum + sums[bot][i]; if(map.containsKey(sum)){ result[0][0] = top; result[0][1] = map.get(sum)+1; result[1][0] = bot; result[1][1] = i; return result; }else{ map.put(sum, i); } } } } return result; } } ``` ####lt944 Maximum Submatrix ``` public class Solution { /** * @param matrix: the given matrix * @return: the largest possible sum */ public int maxSubmatrix(int[][] matrix) { // write your code here if(matrix==null || matrix.length==0 || matrix[0].length==0) return 0; int[][] sums = new int[matrix.length][matrix[0].length]; for(int col=0; col 0 ? sum+sums[bot][i]-sums[top-1][i] : sum+sums[bot][i]; result = Math.max(result, sum-min); min = Math.min(min, sum); } } } return result; } } ``` ####304 Range Sum Query 2D - Immutable ``` class NumMatrix { int[][] sums; public NumMatrix(int[][] matrix) { if(matrix==null || matrix.length==0 || matrix[0]==null || matrix[0].length==0) return; sums = new int[matrix.length+1][matrix[0].length+1]; for(int i=1; i<=matrix.length; i++){ int sum = 0; for(int j=1; j<=matrix[0].length; j++){ sum = sum+matrix[i-1][j-1]; sums[i][j] = sum + sums[i-1][j]; } } } public int sumRegion(int row1, int col1, int row2, int col2) { if(sums==null) return -1; return sums[row2+1][col2+1]-sums[row2+1][col1] - sums[row1][col2+1] + sums[row1][col1]; } } /** * Your NumMatrix object will be instantiated and called as such: * NumMatrix obj = new NumMatrix(matrix); * int param_1 = obj.sumRegion(row1,col1,row2,col2); */ ``` ####308 Range Sum Query 2D - Mutable ``` class NumMatrix { int[][] sums; int[][] matrix; public NumMatrix(int[][] matrix) { if(matrix==null || matrix.length==0 || matrix[0]==null || matrix[0].length==0) return; sums = new int[matrix.length+1][matrix[0].length+1]; this.matrix = matrix; for(int i=1; i<=matrix.length; i++){ int sum = 0; for(int j=1; j<=matrix[0].length; j++){ sum = sum + matrix[i-1][j-1]; sums[i][j] = sums[i-1][j] + sum; } } } public void update(int row, int col, int val) { if(sums==null) return; int delta = val - matrix[row][col]; System.out.println(delta); matrix[row][col] = val; for(int i=row+1; i