剑指 Offer II 095. 最长公共子序列剑指 Offer II 095. 最长公共子序列给定两个字符串 tex

两个字符串各自取任意长度，对应的最优质递推过程，这里使用了二维数组的思维

var longestCommonSubsequence = function(text1, text2) {
    const m = text1.length, n = text2.length;
    const dp = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
    for (let i = 1; i <= m; i++) {
        const c1 = text1[i - 1];
        for (let j = 1; j <= n; j++) {
            const c2 = text2[j - 1];
            if (c1 === c2) {
                dp[i][j] = dp[i - 1][j - 1] + 1;
            } else {
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
    }
    return dp[m][n];
};