[LeetCode][JavaScript]Maximum Subarray

Maximum Subarray

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [−2,1,−3,4,−1,2,1,−5,4],
the contiguous subarray [4,−1,2,1] has the largest sum = 6.

https://leetcode.com/problems/maximum-subarray/

 

 

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找出和最大的子串。

动态规划 ，维护一个变量previous，记录之前的最大值。

当前的最大值就是Math.max(previous + nums[i], nums[i])。

 

/**
 * @param {number[]} nums
 * @return {number}
 */
var maxSubArray = function(nums) {
    if(nums.length === 0) return 0;
    var previous = Math.max(0, nums[0]), max = nums[0];
    for(var i = 1; i < nums.length; i++){
        previous = Math.max(previous + nums[i], nums[i]);
        max = Math.max(previous, max);
    }
    return max;
};

 

 

一开始写的比较啰嗦。

动态规划，到当前index的子串的最大值可能有三种情况：

1. 当前元素的，nums[i]

2. nums[i] + 包括了上一个元素nums[i - 1]的最大子串的和

3. nums[i] + 上一个元素nums[i - 1]之前的最大值(不包括nums[i - 1]) + nums[i - 1]

 

/**
 * @param {number[]} nums
 * @return {number}
 */
var maxSubArray = function(nums) {
    if(nums.length === 0) return 0;
    var dp = [], max = nums[0], previous, current;
    dp[0] = {previous : 0, current: nums[0]};
    for(var i = 1; i < nums.length; i++){
        previous = dp[i - 1].current;
        current = Math.max(nums[i], nums[i] + dp[i - 1].current,
            nums[i] + nums[i - 1] + dp[i - 1].previous);
        dp[i] = {previous : previous, current: current};
        max = Math.max(current, max);
    }
    return max;
};