leetcode 410. Split Array Largest Sum

Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays.Note:If n is the length of array, assume the following constraints are satisfied:1 ≤ n ≤ 10001 ≤ m ≤ min(50, n)E

这题有两种解法，动态规划和二分搜索，动态规划需要三层循环，不如二分搜索高效

DP solution. This is obviously not as good as the binary search solutions; but it did pass OJ.dp[s,j] is the solution for splitting subarray n[j]...n[L-1] into s parts.dp[s+1,i] = min{ max(dp[s,j], n[i]+...+n[j-1]) }, i+1 <= j <= L-sThis solution does not take advantage of the fact that the numbers are non-negative (except to break the inner loop early). That is a loss. (On the other hand, it can be used for the problem containing arbitrary numbers)public int splitArray(int[] nums, int m){    int L = nums.length;    int[] S = new int[L+1];    S[0]=0;    for(int i=0; i<L; i++)        S[i+1] = S[i]+nums[i];    int[] dp = new int[L];    for(int i=0; i<L; i++)        dp[i] = S[L]-S[i];    for(int s=1; s<m; s++)    {        for(int i=0; i<L-s; i++)        {            dp[i]=Integer.MAX_VALUE;            for(int j=i+1; j<=L-s; j++)            {                int t = Math.max(dp[j], S[j]-S[i]);                if(t<=dp[i])                    dp[i]=t;                else                    break;            }        }    }    return dp[0];}

采用二分搜索

public class Solution {    public int splitArray(int[] nums, int m) {        int max = 0; long sum = 0;        for (int num : nums) {            max = Math.max(num, max);            sum += num;        }        if (m == 1) return (int)sum;        //binary search        long l = max; long r = sum;        while (l <= r) {            long mid = (l + r)/ 2;            if (valid(mid, nums, m)) {                r = mid - 1;            } else {                l = mid + 1;            }        }        return (int)l;    }    public boolean valid(long target, int[] nums, int m) {        int count = 1;        long total = 0;        for(int num : nums) {            total += num;            if (total > target) {                total = num;                count++;                if (count > m) {                    return false;                }            }        }        return true;    }}

vaild函数是一种贪心的将数组分成多份的方法，每次累加直到超过target。如果分出的分数大于m，说明target太小了，应该增加，在二分搜索中修正l = mid +1；如果小于m，说明target选的太大了，修正r = mid-1。如果等于m，说明安装贪心是一种分的方法，寻求能否更小，所以修正r = mid-1。